Sorry, I dont think this should be to hard, but Im not sure what this falls under or how to ask best.

So, if I have three numbers, like 5 6 and 7 or 2.02 2.03 and 2.04, and I want to know what new number these three numbers can be devided by so they are all as close to the same value,
so

old weight 1 = similar new weight * same number
old weight 2 = similar new weight * same number
old weight 3 = similar new weight * same number

I also want a balance where the same number is as close to 1 as possible, and where the percent of change in the same number from one is less then the percent of difference between the new similar weights.

Jul 9th 2010, 03:29 AM

Grandad

More information please

Hello knowthebird

Welcome to Math Help Forum!

Quote:

Originally Posted by knowthebird

Sorry, I dont think this should be to hard, but Im not sure what this falls under or how to ask best.

So, if I have three numbers, like 5 6 and 7 or 2.02 2.03 and 2.04, and I want to know what new number these three numbers can be devided by so they are all as close to the same value,
so

old weight 1 = similar new weight * same number
old weight 2 = similar new weight * same number
old weight 3 = similar new weight * same number

I also want a balance where the same number is as close to 1 as possible, and where the percent of change in the same number from one is less then the percent of difference between the new similar weights.

I'm sorry, but you'll have to explain this again in a more precise way. What you've written here makes no real sense at all.

Can you try again, possibly with some complete examples?

Grandad

Jul 9th 2010, 03:56 AM

knowthebird

Yes I can.

So I have three weights, maybe more, maybe only two. The weights i have are equal to a weight times an effectiveness factor of 1. So I do some math to find out how much weight should have been needed to do some amount of work. But the weights are not all the same even though they should be. So if I had the weights 2.01 2.02 and 2.03, this would equal some weight time effectiveness factor of 1. I want a factor close to one that when devided by my three weights will give a new weight for each one that are closer togethor. So for numbers 6 8 and 12, changing the factor to 2 would give weights 3 4 and 6. 3 4 and 6 are closer to the same number so this is better. what I dont want is some huge effectivenes factor like 6, giving 1 1.3 and 2. I want the least amount of change on the effectiveness factor which will bring the most change to the weights bringing them as close togethor as possible.

when i solve all these i get different weights with the same effectiveness factor of 1, i want the weights to be the same for each distance and the effectiveness factor to be the same for each distance, or as close as they can be to the same.

EDIT: I know the same amount of weight was used to do the same amount of work to different objects at different distances. But the weights i get from distances vary, so my line of thought it that a small change in a greater effectiveness factor times some weight could cause a significant change in the distance, and the discrepencies in the distances could lead to an estimate of the effectiveness factor.

Jul 10th 2010, 08:59 AM

ebaines

You need to define what you mean by "as close together as possible." In your example you clain that the numbers 3,4, and 6 are "closer together" than 6, 8, and 12, yet on a percentage basis 3,4, and 6 are as far from each other as 6, 8, and 12 are. The numbers 3,4 and 6 really aren't "closer together" except in the sense that the absolute difference is a smaller number. If this is truly what you;re looking for, then to minimize the difference in your numbers you must divide the original numbers by as large a factor as possible. You are going to have to quantify what you mean by "I want the least amount of change on the effectiveness factor which will bring the most change to the weights bringing them as close togethor as possible." In your example of starting with 6, 8, and 12 - what factor do you think actually works well, and why? Then perhaps we can all understand a bit better just what you're looking for.